HDU1312 深搜 red and black-优快云博客

本文链接：https://blog.youkuaiyun.com/winehownine/article/details/58157649

本文解析了HDU1312题目，该题目要求在一个由黑色和红色方砖组成的矩形房间里，从指定的黑色方砖出发，计算能够到达的所有黑色方砖的数量。文中提供了一种通过深度优先搜索算法来解决该问题的方法。

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hdu1312

Problem Description

There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four adjacent tiles. But he can't move on red tiles, he can move only on black tiles.

Write a program to count the number of black tiles which he can reach by repeating the moves described above.

Input

The input consists of multiple data sets. A data set starts with a line containing two positive integers W and H; W and H are the numbers of tiles in the x- and y- directions, respectively. W and H are not more than 20.

There are H more lines in the data set, each of which includes W characters. Each character represents the color of a tile as follows.

'.' - a black tile
'#' - a red tile
'@' - a man on a black tile(appears exactly once in a data set)

Output

For each data set, your program should output a line which contains the number of tiles he can reach from the initial tile (including itself).

Sample Input

6 9
....#.
.....#
......
......
......
......
......
#@...#
.#..#.
11 9
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
...@...
###.###
..#.#..
..#.#..
0 0

Sample Output

大概题意：在一个迷宫里 @为起点#不能走。能走求能走的最大步数

思路：深度搜索用全局变量求递归函数调用次数即为所求

#include<iostream>

using namespace std;
char room[21][21];
int n,m,ans=0,dx[4]={1,0,-1,0},dy[4]={0,1,0,-1};
void dfs(int x,int y)
{
    ans++;
    int i,xx,yy;
    room[x][y]='#';
    for(i=0;i<4;i++)
    {
        xx=x+dx[i];
        yy=y+dy[i];
        if(xx>=0&&xx<m&&yy<n&&yy>=0&&room[xx][yy]=='.')
        {
            dfs(xx,yy);
        }
    }

}
int main()
{

    int i,j;
    while(cin>>n>>m&&n!=0&&m!=0)
    {
        for(i=0;i<m;i++)
        for(j=0;j<n;j++)
        {
            cin>>room[i][j];
        }
        for(i=0;i<m;i++)
        for(j=0;j<n;j++)
        {
            if(room[i][j]=='@')
            {dfs(i,j);
             break;
            }
        }
        cout<<ans<<endl;
        ans=0;
    }
    return 0;

}